Amphiphilic molecule: __________

C3H8(g) + 5O2(g) ⟶ 3CO2(g) + 4H2O(g) H = -2220 kJ If 865.9 g of H2O is produced during this combustion, how much heat is generat

dem82 [27]

Answer:

3 × 10⁴ kJ

Explanation:

Step 1: Write the balanced thermochemical equation

C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(g) ΔH = -2220 kJ

Step 2: Calculate the moles corresponding to 865.9 g of H₂O

The molar mass of H₂O is 18.02 g/mol.

865.9 g × 1 mol/18.02 g = 48.05 mol

Step 3: Calculate the heat produced when 48.05 moles of H₂O are produced

According to the thermochemical equation, 2220 kJ of heat are evolved when 4 moles of H₂O are produced.

48.05 mol × 2220 kJ/4 mol = 2.667 × 10⁴ kJ ≈ 3 × 10⁴ kJ

5 0

2 years ago

How much heat was transferred from 50.0 g of water if the temperature of the water went from 30.0 ° C to 55.0 °? The specific he

Lena [83]

Answer:

Heat transfer = Q = 62341.6 J

Explanation:

Given data:

Heat transfer = ?

Mass of water = 50.0 g

Initial temperature = 30.0°C

Final temperature = 55.0°C

Specific heat capacity of water = 4.184 J/g.K

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 55.0°C - 30.0°C

ΔT = 25°C (25+273= 298 K)

Q = 50.0 g × 4.184 J/g.K ×298 K

Q = 62341.6 J

4 0

2 years ago

Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an

garri49 [273]

Explanation:

The given data is as follows.

T = $120^{o}C$ = (120 + 273.15)K = 393.15 K,

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So, $x_{1}$ = 0.5 and $x_{2}$ = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

$p^{sat}_{1}$ (393.15 K) = 9.2 bar

$p^{sat}_{1}$ (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

$p_{1} = x_{1} \times p^{sat}_{1}$

= $0.5 \times 9.2 bar$

= 4.6 bar

and, $p_{2} = x_{2} \times p^{sat}_{2}$

= $0.5 \times 10.5 bar$

= 5.25 bar

Therefore, the bubble pressure will be as follows.

P = $p_{1} + p_{2}$

= 4.6 bar + 5.25 bar

= 9.85 bar

Now, we will calculate the vapor composition as follows.

$y_{1} = \frac{p_{1}}{p}$

= $\frac{4.6}{9.85}$

= 0.467

and, $y_{2} = \frac{p_{2}}{p}$

= $\frac{5.25}{9.85}$

= 0.527

Calculate the dew point as follows.

$y_{1}$ = 0.5, $y_{2}$ = 0.5

$\frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}$

$\frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}$

$\frac{1}{P}$ = 0.101966 $bar^{-1}$

P = 9.807

Composition of the liquid phase is $x_{i}$ and its formula is as follows.

$x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{9.2}$

= 0.5329

$x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{10.5}$

= 0.467

4 0

3 years ago

Find the density of an object that has a mass of 12.69 grams and a volume of 3.5cm3

lutik1710 [3]

Answer:

3.62 g/cm³

Explanation:

density = mass ÷ volume

Therefore, do 12.69 divided by 3.5

6 0

2 years ago

The theoretical yield of Cl2 from certain starting amounts of MnO2 and HCl was calculated as 60.25 g and 65.02 g, respectively.

hoa [83]

The actual yield is 43 g Cl₂.

The <em>limiting reactant was MnO₂</em> because it gave the smaller mass of Cl₂.

∴ The <em>theoretical yield</em> is 60.25 g Cl₂.

% yield = actual yield/theoretical yield × 100 %

Actual yield = theoretical yield × (% yield/100 %) = 60.25 g × (72 %/100%) = 43 g

8 0

3 years ago

Amphiphilic molecule: ___________